What is the integral of sqrt(x^8+2x^6+x^4) ?

The integral of sqrt(x^8+2x^6+x^4) is (x^5)/5+(x^3)/3+C

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integral of sqrt(x^8+2x^6+x^4)

\int x^{8} + 2 x^{6} + x^{4} d x

\frac{x ^{5}}{5} + \frac{x ^{3}}{3} + C

Solution steps

\int x^{8} + 2 x^{6} + x^{4} d x

Simplify

x^{8} + 2 x^{6} + x^{4} : x^{2} (x^{2} + 1)

= \int x^{2} (x^{2} + 1) d x

Expand

x^{2} (x^{2} + 1) : x^{4} + x^{2}

= \int x^{4} + x^{2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{4} d x + \int x^{2} d x

\int x^{4} d x = \frac{x ^{5}}{5}

\int x^{2} d x = \frac{x ^{3}}{3}

= \frac{x ^{5}}{5} + \frac{x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{5}}{5} + \frac{x ^{3}}{3} + C

3 t (\frac{d y}{d t}) + y = t^{2}

f (x) = x^{2} + 16

(2 x y + 2 x) d x - (1 + x^{4}) d y = 0

y^{^{'}} + 8 (tan (8 x)) y = 10 cos (8 x)

\int 11 sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

What is the integral of sqrt(x^8+2x^6+x^4) ?
The integral of sqrt(x^8+2x^6+x^4) is (x^5)/5+(x^3)/3+C